Engineering Physics Questions and Answers – Mass Variation with Velocity

«
»

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Mass Variation with Velocity”.

1. The lowest possible mass of a particle is its _____________
a) Relativistic mass
b) Inertial mass
c) Gravitational mass
d) Rest mass
View Answer

Answer: d
Explanation: We know the relation, m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Thus, the minimum velocity one can acquire is zero. For that, the expression reduces to m = m0.
Thus, rest mass is the lowest possible mass of a particle.
advertisement

2. The rest mass of a photon is equal to ___________
a) Gravitational mass
b) Relativistic mass
c) Inertial mass
d) Zero
View Answer

Answer: d
Explanation: A photon is defined as a quantum of energy. It travels with the sped of light and is said to have no rest mass.

3. If the direction of velocity is changed by 180o simultaneously, the mass would _____________
a) Change simultaneously
b) Not Change
c) Increase
d) Decrease
View Answer

Answer: b
Explanation: As we know, the relation for mass variation with velocity is m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\).
As we can see, it depends on the square of the velocity. Thus, changing the direction of velocity would not effect it.

4. Which basic law is used for the derivation of mass variation with velocity?
a) Law of conservation of Energy
b) Law of conservation of Kinetic Energy
c) Law of conservation of Momentum
d) Law of conservation of mass
View Answer

Answer: c
Explanation: The law of conservation of momentum is used for the derivation of the formula. It is assumed that two particles in a frame collides elastically and are rebounded.

5. The speed of light is a limiting velocity, unattainable by a material body.
a) True
b) False
View Answer

Answer: a
Explanation: The velocity of light is the limiting velocity. From the expression, m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\), as the velocity of the object approaches the speed of light, it’s relativistic mass approaches infinity. Thus, the speed of light is unattainable by a material body. Only those with zero rest mass, i.e., photons, can attain the speed of light.
advertisement

6. Which of the following graph shows the correct variation of mass with velocity?
a)engineering-physics-questions-answers-mass-variation-velocity-q6a
b) engineering-physics-questions-answers-mass-variation-velocity-q6b
c) engineering-physics-questions-answers-mass-variation-velocity-q6c
d) engineering-physics-questions-answers-mass-variation-velocity-q6d
View Answer

Answer: c
Explanation: We know the relation m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Thus, according to it, the below graph shows the correct variation of mass with velocity.
engineering-physics-questions-answers-mass-variation-velocity-q6c

7. At what speed the mass of an object would be double its value at rest?
a) √2 c
b) c/√2
c) √3c/2
d) 2c/√3
View Answer

Answer: c
Explanation: Using the relation, m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Here, m = 2m0. Therefore, 2 m0. = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
4(1 – v2/c2) = 1
V = √3 c/2.

8. A man, who weighs 60 kg on earth, weighs 61 kg on a rocket, as measured by an observer on earth. What is the speed of the rocket?
a) 2.5 X 108 m/s
b) 2.5 X 107 m/s
c) 5.5 X 107 m/s
d) 5.5 X 108 m/s
View Answer

Answer: c
Explanation: Now, as we know m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Here, m = 61 kg and m0 = 60 kg
Therefore, 61/60 = 1/√\(\sqrt{1-\frac{v^2}{c^2}}\)
v2/c2 = 1 – (60/61)2 = 121/3600
v = 11 X c/60 = 5.5 X 107 m/s.

9. An object of rest mass 6 Kg is moving with a speed of 0.8c. Its effective mass is ____________
a) 6 Kg
b) 8 Kg
c) 10 Kg
d) 12 Kg
View Answer

Answer: c
Explanation: We know the relation, m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Now, here v = 0.8 c
Therefore, v2/c2 = 0.64
Hence, 1 -v2/c2 = 0.36
Thus, the expression reduces to = 6/0.6 Kg
= 10 Kg.
advertisement

10. A body is initially at rest. It explodes into two objects of mass 1 Kg and 2 Kg moving with a speed of 0.6 times the speed of velocity each. What would be the mass of the original body at rest?
a) 2 Kg
b) 2.5 Kg
c) 3.5 Kg
d) 3.75 Kg
View Answer

Answer: d
Explanation: E0 = mc2 = γm1c2 + γm2c2
E0 = \(\frac{m1C^2}{\sqrt{1-\frac{v^2}{c^2}}} + \frac{m2C^2}{\sqrt{1-\frac{v^2}{c^2}}}\)
m = E/c2
Therefore, m = \(\frac{m1+m2}{\sqrt{1-\frac{v^2}{c^2}}}\)
m = 3/0.8
m = 3.75 Kg.

Sanfoundry Global Education & Learning Series – Engineering Physics.

To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn